Question
✓
Answer
- 120º
Solution:
In $\triangle\text{ABD}$
$\angle\text{A} + \angle\text{B} + \angle\text{D} = 180^\circ$
$⇒ 55^\circ + \angle\text{DBA} + 25^\circ = 180^\circ$
$⇒ \angle\text{DBA} = 180^\circ - 55^\circ - 25^\circ$
$= 180^\circ - 80^\circ$
$⇒ \angle\text{DBA} = 100^\circ$
So, $\angle\text{DBC} = 180^\circ - \angle\text{DBA}$
$= 180^\circ - 100^\circ$
$ \angle\text{DBC} = 80^\circ$
Now, $\triangle\text{EBC}$
$\angle\text{A} + \angle\text{EBC} + \angle\text{C} = 180^\circ$
$⇒ \angle\text{E} + 80^\circ + 40^\circ = 180^\circ ( \angle\text{DBC} = \angle\text{EBC})$
$⇒ \angle\text{E} = 180^\circ - 120^\circ = 60^\circ$
Also, $\text{x} = 180^\circ - \angle\text{E} = 180^\circ - 60^\circ$
$⇒ \text{x} = 120^\circ$
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